Can anyone help?

Question

{ 2+x if x < -1

f (x) ={ [[x-1]] if -1 ≤ x < 1

{ x^2-4x / 4-x if ≥ 1

Find all the possible points of discontinuity of f

Discuss the continuity of f at the points obtained in item 1

Is f continuous on the interval [0,2]? Justify your answer

Answer 1

You are defining a function whose value depends on which interval of x you are on.

The three intervals of "possible input values" are:

x < -1

-1 ≤ x < 1

1 ≤ x Noting that I have written this interval in the reverse direction as originally given, x ≥ 1, to better see the interval flow. I find it helpful to always keep inequalities in the same direction with lesser on left and greater on right, so that they line up with the number line.

For the function to be continuous, it has to be so within each defined interval AND at the transition values of x.

First we note that there is no break in the intervals.

x < -1 AND -1 ≤ x < 1 AND 1 ≤ x has nothing missing for x. There are no breaks between the intervals, there is no discontinuity in the domain of x values. That's a good start.

Then we look at the transition x values of -1 and 1 and ask ourselves if the function value of one interval ends at the same value the next interval starts and includes.

In other words:

Does 2+x=[[x-1]] when x=-1

and does [[x-1]]= x^2-4x / 4-x when x=1

Now, the question does not ask about continuity over the full domain (all possible x values), but rather over the interval of [0,2]={0≤x≤ 2}

Therefore, we are only concerned about intervals 0≤x≤ 1 and 1≤x≤ 2 rather than -1≤x≤ 1 and 1≤x respectively.

Since the first transition of x=-1 is less than 0, we only have to concern ourselves with the second transitional x value, x=1.

NOTE that the transitional x value, x=1, is included in at least one of the two intervals that end and start at x=1. If that were not the case, i.e. if neither interval included x=1, then

x=1 would have been a discontinuity WITHOUT looking at the function definitions over each interval.

At the x transitional value, x=1

For the lesser interval, [[x-1]]= the greatest integer less than or equal to x-1

At x= 1, x-1= 1-1= 0 and the greatest integer less than or equal to 0 is 0

AND for the greater interval (following proper orders of operations)

At x= 1, x^2-4x / 4-x = 1^2-4*1 / 4-1 =

1-4*1 / 4-1 = 1-4/4-1 = 1-4/4-1= 1-1-1= -1

SO, x=1 is a discontinuity because the value of the function over 0 ≤ x < 1 ends with a different value than the function continues with over 1 ≤ x.

THEREFORE; The answer to the question at hand is NO. The function is NOT continuous over [0,2], because if the discontinuity at the transition x=1.